1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Extending a uniformly cont function on an open interval to a closed interval?

  1. Oct 23, 2011 #1
    1. The problem statement, all variables and given/known data
    Show that the function [itex]f: J → ℝ[/itex] is bounded if [itex]f[/itex] is uniformly continuous on the bounded interval J.


    2. Relevant equations
    J is a bounded interval, so say J = (a,b)

    [itex]f[/itex] is uniformly continuous on J, so
    [itex]\forall \epsilon > 0[/itex] there exists a [itex]\delta > 0[/itex] such that for [itex]s,t \in J = (a,b)[/itex]
    [itex]|f(s) - f(t)| < \epsilon[/itex] whenever [itex]|s - t| < \delta[/itex]

    [itex]f: J → ℝ[/itex] is bounded if there exists a real number M such that |f(x)| ≤ M for all x in J.


    3. The attempt at a solution
    I think that if I can extend [itex]f[/itex] to the endpoints of J, then I can use the Extreme Value Theorem to say that [itex]f[/itex] attains a min and max value, i.e. is bounded. So I need to define [itex]f(a)[/itex] and [itex]f(b)[/itex].

    [itex]f(a) = \lim_{n→∞}{f(a + \frac{1}{n})}[/itex]
    [itex]f(b) = \lim_{n→∞}{f(b - \frac{1}{n})}[/itex]

    Where clearly [itex]f[/itex] is defined on [itex][a + \frac{1}{n}, b - \frac{1}{n}][/itex] for all [itex]n \in ℕ[/itex]


    Am I on the right track? I don't feel like I've used the fact that [itex]f[/itex] is uniformly continuous on J. Is it because [itex]f[/itex] is uniformly continuous that I'm able to define [itex]f(a)[/itex] and [itex]f(b)[/itex]?
     
  2. jcsd
  3. Oct 23, 2011 #2
    Try to do this more simply. The idea is to pick x1,....,xn such that every x in J is "close" to one of these xi. This plus uniform continuity will give you what you want.
     
  4. Oct 23, 2011 #3
    But I only know that J is bounded, not that it's closed and bounded. So can't I only say that J = (a,b) and not necessarily J = [a,b]? So I don't immediately know if f is defined at a or b, much less if it's continuous at a and b.
     
  5. Oct 23, 2011 #4
    Okay, nevermind, you changed your post. So... I should define a sequence of, say, rational numbers {xn}, where all xi are in J, such that [itex]lim_{n->∞}{\{x_n\}} = a[/itex] ?

    And similarly, {yn}, where all yi are in J, such that [itex]lim_{n->∞}{\{y_n\}} = b[/itex] ?

    I think that I'm able to do this, but I don't see how it gets me any closer to showing that f is bounded.

    EDIT:

    I guess I can say that

    [itex]lim_{n->∞}{f(x_n)} = f(a)[/itex]
    [itex]lim_{n->∞}{f(y_n)} = f(b)[/itex]

    So f is defined on [a,b]. Do I still have to show it's continuous at a and b to use the extreme value theorem? I don't see how this approach is simpler or much different than the one in my original post.
     
    Last edited: Oct 23, 2011
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Extending a uniformly cont function on an open interval to a closed interval?
Loading...